A finitely generated flat $A$-module $M$ is faithfully flat if and only if $\operatorname{Ann}(M)=0$

How one can show that a finitely generated flat $A$-module $M$ is faithfully flat if and only if $\operatorname{Ann}(M)=0$? (Liu, Algebraic Geometry and Arithmetic Curves, Exercise 2.17.)

I tried to show that the condition $N$ is an $A$-module with $M\otimes_AN=0$ then $N=0$ implies that $\{a\in A:aM=0\}=\{0\}$ but I didn't find the proof.

Let $\mathfrak a=\operatorname{Ann}_A (M)$.
If part: we have to prove that for any maximal ideal $\mathfrak m$ of $A$, $M\neq\mathfrak m M$. Since $M$ is finitely generated, $\operatorname{Supp}(M)= V(\mathfrak a)=\operatorname{Spec}A$. Hence $M_{\mathfrak m}\neq 0$. Now if $M=\mathfrak m M$ were true, then localising at $\mathfrak m$, we'd have $M_{\mathfrak m}=\mathfrak m M_{\mathfrak m}$ and, by Nakayama's lemma, $M_{\mathfrak m}=0$.
Only if part: consider the canonical injection $i\colon\mathfrak a \hookrightarrow A$. Tensoring with $M$, $i\otimes_A \operatorname{id}_M$ remains injective (flatness); however it is $0$ since $\mathfrak a$ is the annihilator of $M$. Hence $\mathfrak a\otimes_A M=0$, so that $\mathfrak a=0$ (faithful flatness).